All categories

3 years ago

Find the equation of the circle with the given characteristics. Center: (−2, 1) Radius: 2 The solution is

1 answer:

8 0

The format for the equation of a circle is (x-h)^2+(y-k)^2=r^2, where (h,k) is your center and r is your radius. All we have to do is substitute the correct values, giving us the equation (x+2)^2+(y-1)^2=4

:)

You might be interested in

A coin is biased to show 35% heads and 65% tails. the coin is tissed twice. what is the probability that the coin turns up heads

bekas [8.4K]

The required probability is:
$P(1\ heads\ and\ 1\ tails)=2C1\times0.35\times0.65=0.455$
The answer is 0.455.

6 0

3 years ago

375,000 in scientific notation

Luba_88 [7]

3.75X10^5 That seems to be the answer. You could type that in your calculator to see what you get.

4 0

2 years ago

Find the zeros of the polynomial x2-7<br><br>please answer step by step

Yuki888 [10]

Answer:

x = ± $\sqrt{7}$

Step-by-step explanation:

To find the zeros equate the polynomial to zero, that is

x² - 7 = 0 ( add 7 to both sides )

x² = 7 ( take the square root of both sides )

x = ± $\sqrt{7}$

Thus the exact solutions are

x = - $\sqrt{7}$ , x = $\sqrt{7}$

6 0

2 years ago

Write the vertex form equivalent of 3x^2-30x+73=y

Naddik [55]

Answer:

(5,-2) or y=3(x-5)^2 -2

Step-by-step explanation:

5 0

3 years ago

Somebody please assist me here

Anettt [7]

The base case of $n=1$ is trivially true, since

$\displaystyle P\left(\bigcup_{i=1}^1 E_i\right) = P(E_1) = \sum_{i=1}^1 P(E_i)$

but I think the case of $n=2$ may be a bit more convincing in this role. We have by the inclusion/exclusion principle

$\displaystyle P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1 \cup E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le P(E_1) + P(E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le \sum_{i=1}^2 P(E_i)$

with equality if $E_1\cap E_2=\emptyset$ .

Now assume the case of $n=k$ is true, that

$\displaystyle P\left(\bigcup_{i=1}^k E_i\right) \le \sum_{i=1}^k P(E_i)$

We want to use this to prove the claim for $n=k+1$ , that

$\displaystyle P\left(\bigcup_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)$

The I/EP tells us

$\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cup E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right)$

and by the same argument as in the $n=2$ case, this leads to

$\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1})$

By the induction hypothesis, we have an upper bound for the probability of the union of the $E_1$ through $E_k$ . The result follows.

$\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^k P(E_i) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)$

5 0

2 years ago